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The solution of the differential equation `(1+xy)xdy+(1-xy)ydx=0` ,isA. `(1)/(xy)+log((y)/(x))=C`B. `-xy+log((y)/(x))=C`C. `-(1)/(xy)+log((y)/(x))=C`D. `-(1)/(xy)+log((x)/(y))=C` |
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Answer» Correct Answer - C The given equation can be written as `(xdy+ydx)+xy(xdy-ydx)=0` `rArr" "d(xy)+xy(xdy-ydx)=0` `rArr" "(d(xy))/((xy)^(2))+(xdy-ydx)/(xy)=0` `rArr" "(d(xy))/((xy)^(2))+(dy)/(y)-(dx)/(x)=0` `rArr" "(d(x))/((xy)^(2))+d(logy)-d(logx)=0` `rArr" "(d(xy))/((xy)^(2))+d(logy-logx)=0` `rArr" "(d(xy))/((xy)^(2))+dlog((y)/(x))=0` On integrating, we get `-(1)/(xy)+log((y)/(x))=C` as the required solution. |
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